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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Integrate the rational function in exercise:
$\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}$
✓

Answer

$\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{\text{Ax}+\text{B}}{\text{x}^2+1}+\frac{\text{C}}{\text{x}-1}\dots(\text{i)}$
$\Rightarrow x = (Ax + B) (x - 1) + C(x^2 + 1)$
$\Rightarrow x = Ax^2 - Ax + Bx - B + Cx^2 + C$
Comparing coefficients of $x^2, A + C = 0 ....(ii)$
Comparing coefficients of $x, –A + B = 1 ....(iii)$
Comparing constant terms, $–B + C = 0 ….(iv)$
Solving eq. $(ii), (iii)$ and $(iv)$, we get
$\text{A}=\frac{-1}{2},\ \text{B}=\frac{1}{2} \ \text{and} \ \text{C}=\frac{1}{2}$
Putting the values of $A, B$ and $C$ in eq.$(i),$
$\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{\frac{-1}{2}\text{x}+\frac{1}{2}}{\text{x}^2+1}+\frac{\frac{1}{2}}{\text{x}-1}$
$\Rightarrow \ \frac{\text{x}}{(\text{x}^2+1)}=\frac{-1}{2}.\frac{\text{x}}{\text{x}^2+1}+\frac{1}{2}.\frac{1}{\text{x}^2+1}+\frac{1}{2}.\frac{1}{\text{x}-1}=\frac{-1}{4}.\frac{2\text{x}}{\text{x}^2+1}+\frac{1}{2}.\frac{1}{\text{x}^2+1}+\frac{1}{2}.\frac{1}{\text{x}-1}$
$\Rightarrow \ \frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}\text{dx}=\frac{-1}{4}\int\frac{2\text{x}}{\text{x}^2+1}\text{dx}+\frac{1}{2}\int\frac{1}{\text{x}^2+1}\text{dx}+\frac{1}{2}\int\frac{1}{\text{x}-1}\text{dx}$
$​​​​​​\Rightarrow \ \frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}\text{dx}=\frac{-1}{4}\text{log}|\text{x}^2+1|+\frac{1}{2}\tan^{-1}\text{x}+\frac{1}{2}\text{log}|\text{x}-1|+\text{c}$
$​​​​​​\Rightarrow \ \frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{-1}{4}\text{log}(\text{x}^2+1)+\frac{1}{2}\tan^{-1}\text{x}+\frac{1}{2}\text{log}|\text{x}-1|+\text{c}$

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